Problems on Isosceles Triangles with Detailed Solutions

Problems on isosceles triangles are presented along with their detailed solutions.

## Isosceles Triangle Formulas

An Isosceles triangle has two equal sides with the angles opposite to them equal. The relationship between the lateral side $$a$$, the based $$b$$ of the isosceles triangle, its area A, height h, inscribed and circumscribed radii r and R respectively are give by:   ### Problem 1

What is the area of an isosceles triangle with base b of 8 cm and lateral a side 5 cm?

### Problem 2

What is the base of an isosceles triangle with lateral side a = 5 cm and area 6 cm 2?

### Problem 3

What is the lateral side of an isosceles triangle with area 20 unit 2 and base 10 units?

### Problem 4

What is the lateral side of an isosceles triangle such that its height h ( perpendicular to its base b) is 4 cm shorter than its base b and its area is 30 cm 2?

### Problem 5

ABC and BCD are isosceles triangles. Find the size of angle BDE. ### Problem 6

ABC and CDE are isosceles triangles. Find the size of angle CED. ### Problem 7

Find the area of the circle inscribed to an isosceles triangle of base 10 units and lateral side 12 units.

### Problem 8

Find the ratio of the radii of the circumscribed and inscribed circles to an isosceles triangle of base b units and lateral side a units such that a = 2 b.

### Problem 9

Find the lateral side and base of an isosceles triangle whose height ( perpendicular to the base ) is 16 cm and the radius of its circumscribed circle is 9 cm.

### Problem 10

What is the area of an isosceles triangle of lateral side 2 units that is similar to another isosceles triangle of lateral side 10 units and base 12 units?

## Solutions to the Above Questions

1. Solution

Apply Pythagora’s theorem to the right triangle CC’B (see figure at top) to write

a 2 = (b/2) 2 + h 2
h = a 2 – (b/2) 2 = 5 2 – 4 2 which gives h = 3

Area A = (1/2) b h = (1/2) 8�3 = 12 unit 2
2. Solution

Use formula of area of isosceles triangle

A = (1/2) a 2 sin(α)

to find α as follows

sin(α) = 2 A / a 2 = 2 * 6 / 5 2 = 12 / 25

α = arcsin(12 / 25)

Using right triangle CC’B (see figure at top), we can write

sin(α/2) = (b/2) / a = b / 2a

b = 2 a sin (α/2)

b = 2 a sin( (1/2) arcsin(12 / 25) ) = 10 sin( (1/2) arcsin(12 / 25) ) ≅ 2.48
3. Solution

Use formula of area of isosceles triangle to write

A = (1/2) b h = 20

Given b = 10, find h

h = 40 / 10 = 4

Pythagora’s theorem used in the right triangle CC’B (see figure at top) to write

a 2 = (b/2) 2 + h 2 = √ ( 5 2 + 4 2) = √41
4. Solution

Use formula of area of isosceles triangle to write

A = (1/2) b h = 30

h = b – 4 (given)

Substitute h by b – 4 in the formula for A

A = (1/2) b (b – 4) = 30

Gives the equation: b 2 – 4 b – 60 = 0

Solutions to the equation: b = 10 and b = – 6

b is a length and therefore is positive b = 10 , h = b – 4 = 6

a = √ (h 2 + (b/2) 2) = √ (36 + 25) = √61
5. Solution

ABC is an isosceles triangle and therefore

∠CAB = ∠ABC

Also: ∠CAB + ∠ABC = 180 – 66

2 ∠ABC = 180 – 66

gives: ∠ABC = 57�

BCD is a right isosceles triangle; hence

∠CBD = ∠CDB = (180 – 90)/2 = 45�

Note that ∠ABC, ∠CBD and ∠DBE make a straight angle. Hence

∠ABC + ∠CBD + ∠DBE = 180�

gives ∠DBE = 180 – ∠ABC – ∠CBD = 180 – 57 – 45 = 78�

DBE is a right triangle ; hence

∠BDE = 90 – 78 = 12�
6. Solution

Straight angle at B hence

∠ABC = 180 – 116 = 64�

ABC is an isosceles triangle and therefore

∠CAB = ∠ABC = 64�

In trangle ABC: ∠BCA + ∠CAB + ∠ABC = 180�

Hence: ∠BCA = 180 – 64 – 64 = 52�

∠DCE = ∠ BCA = 52�

CDE isosceles: ∠DCE = ∠CDE = 52�

In triangle CDE we have: ∠DCE + ∠CDE + ∠CED = 180�

∠CED = 180 – 52 – 52 = 76�
7. Solution

Radius of inscribed circle to an isosceles triangle of base b = 10 and lateral side a = 12 is given by

r = $$\dfrac{b}{2} \sqrt{\dfrac{2a-b}{2a+b}} = 5 \sqrt{\dfrac{4}{34}}$$

Area A of circle of radius r is given by: A = π r 2 = 100π / 34
8. Solution

Radius R of the circumscibed circle and radius r of the inscribed circle to the same isosceles triangle of base b and lateral side a are given by

R = $$\dfrac{a^2}{\sqrt{4a^2-b^2}}$$ and r = $$\dfrac{b}{2} \sqrt{\dfrac{2a-b}{2a+b}}$$

Substitute a by 2 b (given) in both formulas and simplify

R = $$\dfrac{(2b)^2}{\sqrt{4(2b)^2-b^2}} = \dfrac{4b^2}{\sqrt{15b^2}} =\dfrac{4b}{\sqrt15}$$

r = $$\dfrac{b}{2} \sqrt{\dfrac{4b-b}{4b+b}} = (b/2) \sqrt{\dfrac{3}{5}}$$

Simplify ratio: R / r = 8/3
9. Solution

Radius R of the circumscibed circle an isosceles triangle of base b and lateral side a are given by

R = $$\dfrac{a^2}{\sqrt{4a^2-b^2}}$$

Relationship between h, b and a (Pythagora in triangle CC’B see figure above)

a 2 = (b/2) 2 + h 2
Gives: a 2 = b 2 / 4 + h 2 or 4 h 2 = 4 a 2 – b 2
Substitute R by 9 (given) and 4 a 2 – b 2 by 4 h 2 in the formula for R above to get the equation

9 = $$\dfrac{a^2}{\sqrt{4h^2}} = a^2 / (2h)$$

a 2 = 9 � 2 � 16

gives a = 12 √2 cm.
10. Solution

Let A1 and A2 be the areas of the triangle with lateral sides a1 = 2 and a2 = 10 respectively.

Formulas for A1 and A2
A1 = (1 / 2) a1 2sin(α) and A2 = (1 / 2) a2 2 sin(α) , corresponding angles are equal in similar triangles.

Hence the ratio: A1 / A2 = (1 / 2) a1 2sin(α) / (1 / 2)a2 2sin(α) = a1 2 / a2 2
We now need to find A2
Use Pythagora’s theorem in the right triangle CC’B (see figure at top) to write

a 2 = (b/2) 2 + h 2
h = a 2 – (b/2) 2 = 10 2 – 6 2 which gives h = 8

Area A2 = (1 / 2) b h = (1 / 2) 12�8 = 48 unit 2
a1 2 / a2 2 = A1 / A2
A1 = A2 � a1 2 / a2 2 = 48 � 2 2/ 10 2 = 1.92 unit 2

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